Ngân hàng bài tập
B
Tìm đạo hàm của hàm số $$y=\left(x^2-x+1\right)^{\tfrac{1}{3}}$$
 | \(y'=\dfrac{2x-1}{\sqrt[3]{\left(x^2-x+1\right)^2}}\) |
 | \(y'=\dfrac{1}{3\sqrt[3]{\left(x^2-x+1\right)^2}}\) |
 | \(y'=\dfrac{2x-1}{3\sqrt[3]{x^2-x+1}}\) |
 | \(y'=\dfrac{2x-1}{3\sqrt[3]{\left(x^2-x+1\right)^2}}\) |
1 lời giải
Chọn phương án D.
\(\begin{aligned}
y'&=\dfrac{1}{3}\cdot\left(x^2-x+1\right)^{\tfrac{1}{3}-1}\cdot\left(x^2-x+1\right)'\\
&=\dfrac{1}{3}\cdot\left(x^2-x+1\right)^{-\tfrac{2}{3}}\cdot(2x-1)\\
&=\dfrac{2x-1}{3\left(x^2-x+1\right)^{\tfrac{2}{3}}}\\
&=\dfrac{2x-1}{3\sqrt[3]{\left(x^2-x+1\right)^2}}.
\end{aligned}\)
$$\left(u^n\right)'=n\cdot u^{n-1}\cdot u'$$